`K_(1)` and `K_(2)` for dissociation of `H_(2)S ` are `1.0xx10^(-7)` and `1xx10^(-14)`. Calculate `[H^(+)], [HS^(-)], [S^(2-)]` and `[H_(2)S]` in `0.1M H_(2)S` solution. Also report `[H^(+)]` and pH and `K_(a)` for `H_(2)ShArr2H^(+)+S^(2-)`.\

Explanation

`H_(2)S hArr H^(+)+HS^(-)`
`K_(1)=([H^(+)][HS^(-)])/(H_(2)S]=1.0xx10^(-7)`
`.: [H^(+)]=Calpha, [HS^(-)]=Calpha, [H_(2)S]=C(1-alpha)`
or `1.0xx10^(-7)=(Calpha.Calpha)/(C(1-alpha))=(Calpha^(2))/((1-alpha))`
or `1.0xx10^(-7)=(0.1xxalpha^(2))/((1-alpha)) , (.: 1-alpha=~~1)`
`:. alpha=10^(-3)`
`:. [H^(+)]=Calpha =0.1xx10^(-3)=10^(-4)M`
`:. pH=4`
`[HS^(-)]=Calpha=0.1xx10^(-3)=10^(-4)M`
`[H_(2)S]=C(1-alpha)=0.1(1-10^(-3))=10^(-1)M`
Now `HS^(-)` further dissociate to `H^(+)` and `S^(2-)`.
`{:(HS^(-),hArrH^(+),+S^(2-)),((10^(-4)-y),(10^(-4)+y),y):}`
`:. K_(a_(2)=10^(-14))=((10^(-4)+y).y)/((10^(-4)-y))`
Since dissociation of `HS^(-)` in pressence of `H^(+)` will be suppressed.
Therefore `10^(-4)+y=10^(-4)` and `10^(-4)-y=y`.
`:. y= 10^(14)` or `[S^(2-)]=10^(-14)`